Power Problem 5 in Closed Circuits

Since the resistance r1 is unchanged, when the current i is the largest in the circuit, the maximum power is obtained on p1, and it can be seen that ohm's law i = e (r1+r2+r) of a closed circuit, and when r2 = 0, i is maximum, p1 = (e r1+r) 2r1 = 1w

The power is maximum when the external circuit resistance = the internal circuit resistance.

R1 can be regarded as a part of the internal resistance of the power supply, and the internal resistance of the power supply R'=2 ohms when R1 is connected in parallel with R', so when R2=R', P2=2W

Because r1>r r2=0, r1 has the maximum power p1m= [e (r+r1)] 2*r1 = [4 (1+4)] 2*4 =

From R2, the loop resistance is R+R1 = 5 When R2=5, the maximum power P2M = E 2 4R2 = 4 2 4*5 =

As shown in the figure, it is known that the power supply electromotive force e=4V, the internal resistance r=4 ohms, the fixed resistance r1=4 ohms, the maximum resistance of the sliding rheostat R2 is 16 ohms, when S is closed, change the maximum resistance of R2, when R2 takes what value?

1。R1 is the maximum power, the maximum? r2=

2。R2 power maximum, maximum? r2=8 p2=2w

As can be seen from the figure, when the current is the output power of the power supply is the largest, here the internal friction.

The power is equal to the output power of 9W, the efficiency of the power supply is 50%, and the total power of the power supply is 18W.

for 12V; Therefore d is correct;

From p=i2r, it can be seen that the internal resistance of the power supply is 4, because the answer and the sum cannot be determined whether the external circuit is a pure resistance circuit.

Therefore, it is impossible to find the resistance and take the size; Therefore c is wrong;

From p=i2r, it can be obtained that with the increase of current, the internal power consumption increases, so a is wrong;

With the decrease of current, the efficiency of the power supply increases, and the maximum can be close to 100%; Therefore berror;

Therefore, d

When the motor does not rotate, the coil only heats up, Ohm's law applies, and the resistance of the motor coil.

r=u/i=

The electrical power of the motor when it is working normally.

p=u'i'=9v×

The power lost by the motor coil in heat generation.

P loss = i'²r=(

The power output of the electric motor.

P machine = p - p loss =

Electric motor efficiency.

p machine p=

1. Disconnect R2 from the circuit, and the terminal voltage at the disconnection is U=ER1 (R1+R).

2. When the voltage source fails (short circuit), the internal resistance of the network is: r=r1 r=rr1 (r+r1).

The above is using Thevenin's theorem, the equivalent circuit can be derived.

When the equivalent internal resistance r = external resistance r2, the maximum power can be obtained on r2. That is, when R2=R=Rr1 (R+R1), R2 can obtain the maximum power, and the maximum power is Pmax=U (4R)=E R1 (4R+4Rr1).

Total resistance: r total = (r1*r2) (r1+r2)+r

R2 voltage: V2=[E*(R1*R2) (R1+R2)] [(R1*R2) (R1+R2)+R].

When V2 is maximum, R2 is the most powerful.

When the load resistance is 0, the current is maximum, but the voltage on it is 0, so the power is 0 If the load resistance is the largest, the voltage on it can be very large, but because the current is very small, the power is also very small. To maximize the power, the product of voltage and current must be maximized, so the maximum power occurs in the case of Zen Shixiang where the load resistance is equal to the internal resistance of the power supply.

Power p = ui 2

i=u/2rp=

The current is maximum when the load resistance is 0, but the voltage on it is 0, so the power is 0. If the load resistance is large, the voltage on it can be large, but the current is small, so the power is also small. To maximize the power, the product of voltage and current must be maximized, so the maximum power occurs when the load resistance is equal to the internal resistance of the power supply.

Power p=ui2 i=u 2rp=

The load resistance is equal to the internal resistance of the power supply, which is the maximum output power of the power supply. p=/4r

1. You can think of the rheostat as the internal resistance of the power supply.

When the resistance value of the rheostat plus the internal resistance of the power supply = r0, the output power of the power supply is the largest, that is, the power consumed by r0 is the largest.

When the rheostat value is in ohms, R0 consumes the most power.

2. Again, you can think of R0 as the internal resistance of the power supply.

When the resistance value of the rheostat = r0 plus the internal resistance of the power supply, the output power of the power supply is the largest, that is, the power consumed by the rheostat is the largest.

When the value of the rheostat is in ohms, the rheostat consumes the most power.

3. I don't know if the conditions of your topic have changed.

If the original conditions are not considered, the output power of the power supply is maximum when the resistance of the external circuit is equal to the ohm.

p = iir = 2*2* = 2 watts.

Are R0 and rheostat connected in parallel or series?

By p=i 2r=e 2r (r+r) 2=e 2r [(r-r) 2+4rr] when r=r, the output work reed collapse rate is up to 1 pair.

The gear of the power supply is only efficient =iu ie=r (r+r)=1 (1+r r) r large high efficiency 2 wrong.